Alcabitius

Alcabitius, or Al-Quabisi, lived about 980 CE but the system that bears his name was already known about 500 CE. At that time it was already used by Rhetorius [Knappich p.23-25, Koch/Knappich p. 78, Duval 1984, pp. 23-26].
Gorter wrongly mentions that the system is the same as that of Albategnius [Gorter, p. 57].

The house system of Alcabitius for a long time has been the system that was applied by most of the astrologers. From about 500 up to 1500 CE it was so common that it received the name “Standard method” [North, p. 4].
The system has some resemblance with the Porphyri system. Just like Porphyri, Alcabitius divides the quadrants in three equal parts. But, in stead of ecliptical longitude, he uses right ascension. So his system is not based on the ecliptic but on the equator.

The origin of the equator is the axial rotation of the earth: the daily cycle. The Alcabitius system for the first time is not based on the yearly cycle of the ecliptic, but on the daily cycle. This seems to be logical and therefore it is no surprise that this system was leading for such a long time. Alcabitius abandons the idea that houses are only a redivision of the zodiac itself; instead the houses are now based on a separate circle that is specific for the daily motion of the earth.
Because Alcabitius trisects the equatorial quadrants, just like Porphyrius does in longitude, the calculations are based on unequal parts of the equator.

You can also define Alcabitius with daily and nocturnal semi-arcs. If you do so, you divide the time the degree on the ascendant needs to become MC in three equal parts. From he resulting sidereal times you calculate the longitudes. You apply these calculations to the cusps 11 and 12. You can use the same approach for the degree on the ascendant, using the time it needed to reach the horizon, starting from the IC. You can do so only for the eastern hemisphere, the same calculation for the western hemisphere would give wrong results.
Some authors, including myself [Kampherbeek] and Holden [Holden pp. 89-91], have presented the semi-arc construction as the Alcabitius system. The correct approach would be trisecting the quadrants of the heavenly equator.
Houlding does mention the trisection in time but also mentions that the construction itself is a trisection of the equator. [Houlding,p. 105].

Calculation

You can calculate the trisection of the quadrants in the following way:

define the right ascension (RA) of the MC (Sidereal time * 15).

define the RA of the ascendant (convert longitude to RA).

calculate the difference in RA between MC and ascendant and divide the result by 3.

add the result of the previous step to the RA of the MC, this is the RA of cusp 11

add the same value again and the result will be the RA of cusp 12

repeat this for the other quadrants. For the IC you use the RA of the MC plus or minus 180 degrees, for the descendant the RA of the ascendant plus or minus 180 degrees.

An example

Our location again is Enschede (52º 13′ N and 6º 54′ E). Date and time November 2, 2016 (Gregorian calendar), 21:17:30 UT. The resulting sidereal time is 0:35:23.6 (decimal 0,5899018653) and the angle of the ecliptic E is 23° 26′ 13.56586091” (decimal 23,437101628).

MC: 9.62989868323 converted to degrees and minutes: 9°37′48″ Aries
Asc: 123.507983345667 = 3°30’28” Leo
The RA of the MC is siderel times * 15 , in decimals this is:

0.5899018653 . 15 = 8.8485279795

You can calculate the right ascension easily if you know the longitude and the angle of the ecliptic. The formula is:

tan RA = tan L . cos E

RA is right ascension, L stands for longitude and E for the angle of the ecliptic.
You can use this formula if the point you want to calculate does not have latitude, which is always the case for cusps.

This value is in the wrong quadrant, therefore you need to add 180 degrees

-54.184965095604 + 180 = 125.815034904396

The difference between the RA of the MC and the ascendant

125.815034904396 - 8.8485279795 = 116.966506924896

The result defines the size of the fourth quadrant in right ascension. The same value is valid for the opposite second quadrant.

1/3 of this result

116.966506924896 / 3 = 38.988835641632

This is the size of a house in the fourth or in the second quadrant, measured in right ascension.

Add this value to the RA of the MC:

8.8485279795 + 38.988835641632 = 47.837363621132

This is the right ascension of cusp 11.

Again add the same value to the RA of cusp 11:

47.837363621132 + 38.988835641632 = 86.826199262764

This is the right ascension of cusp 12.

If you add the same value again you will get hte RA of the ascendant, but this is not necessary as you already know this value.

You can calculate cusps 2 and three in the same way. Take the RA of the IC (a difference of 180º with the RA of the MC), define the difference with the RA of the ascendant and trisect this. Add the result to the RA of the ascendant to get the RA for cusp 2 and again for cusp 3.

But you can also use a little trick:
add 60 degrees to the RA of cusp 12, this is the RA of cusp 2
add 120 degrees tot he RA of cusp 11, this is the RA of cusp 3

You can easily understand this if you realize that the three parts of quadrant 4 and the three parts of quadrant 1, will always be 180º if combined.